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Chapter 5: Q5.3-4E (page 267)

In Exercises \({\bf{3}}\) and \({\bf{4}}\), use the factorization \(A = PD{P^{ - {\bf{1}}}}\) to compute \({A^k}\) where \(k\) represents an arbitrary positive integer.

4. \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\)

Short Answer

Expert verified

The required answer is \({A^k} = \left( {\begin{array}{*{20}{c}}{4 - 3\left( {{2^k}} \right)}&{ - 12 + 3\left( {{2^{k + 2}}} \right)}\\{1 - {2^k}}&{ - 3 + {2^{k + 2}}}\end{array}} \right)\).

Step by step solution

01

Write the Diagonalization Theorem

The Diagonalization Theorem:An \(n \times n\) matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors. As \(A = PD{P^{ - 1}}\) which has \(D\) a diagonal matrix if and only if the columns of \(P\) are \(n\) linearly independent eigenvectors of \(A\).

02

Find the inverse of the invertible matrix

Consider the given equation form \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\).

As it is given that \(A = PD{P^{ - 1}}\)than by using the formula for \({n^{th}}\) power we get:

\({A^n} = P{D^n}{P^{ - 1}}\).

Compare the given equation form with \(A = PD{P^{ - 1}}\).

\[\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\]

Therefore,

\[\begin{array}{c}A = \left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right)\\P = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\\D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\\{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\end{array}\]

03

Find \({A^k}\)

\[\begin{array}{c}{A^k} = P{D^k}{P^{ - 1}}\\ = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{2^k}}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( {{2^k}} \right) + 0}&{0 + 4}\\{{2^k} + 0}&{0 + 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( {{2^k}} \right)}&4\\{{2^k}}&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 3\left( {{2^k}} \right) + 4}&{3\left( {{2^k}} \right)\left( 4 \right) - 12}\\{ - {2^k} + 1}&{4\left( {{2^k}} \right) - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{4 - 3\left( {{2^k}} \right)}&{ - 12 + 3\left( {{2^{k + 2}}} \right)}\\{1 - {2^k}}&{ - 3 + {2^{k + 2}}}\end{array}} \right)\end{array}\]

Thus, \({A^k} = \left( {\begin{array}{*{20}{c}}{4 - 3\left( {{2^k}} \right)}&{ - 12 + 3\left( {{2^{k + 2}}} \right)}\\{1 - {2^k}}&{ - 3 + {2^{k + 2}}}\end{array}} \right)\).

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Most popular questions from this chapter

Let \(A\) be a real \(2 \times 2\) matrix with a complex eigenvalue \(\lambda = a - bi\)\(\left( {b \ne 0} \right)\) and an associated eigenvector \({\bf{v}}\) in \({\mathbb{}^2}\).

  1. Show that \(A({\mathop{\rm Re}\nolimits} {\bf{v}}) = a{\mathop{\rm Re}\nolimits} {\bf{v}} + b{\mathop{\rm Im}\nolimits} {\bf{v}}\) and \(A({\mathop{\rm Im}\nolimits} {\bf{v}}) = - b{\mathop{\rm Re}\nolimits} {\bf{v}} + a{\mathop{\rm Im}\nolimits} {\bf{v}}\). (Hint: Write \({\bf{v}} = {\mathop{\rm Re}\nolimits} {\bf{v}} + i{\mathop{\rm Im}\nolimits} {\bf{v}}\), and compute \(A{\bf{v}}\).)
  2. Verify that if \(P\) and \(C\) are given as in Theorem 9, then \(AP = PC\)

In Exercises 9鈥�16, find a basis for the eigenspace corresponding to each listed eigenvalue.

10. \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), \(\lambda = 4\)

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

13. \(\left[ {\begin{array}{*{20}{c}}6&- 2&0\\- 2&9&0\\5&8&3\end{array}} \right]\)

Question: A is a \({\bf{7}} \times {\bf{7}}\) matrix with three eigenvalues. One eigenspace is two-dimensional and one of the other eigenspaces is three-dimensional. Is it possible that A is not diagonalizable? Justify your answer.

Question: Is \(\lambda = 4\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}3&0&{ - 1}\\2&3&1\\{ - 3}&4&5\end{array}} \right)\)? If so, find one corresponding eigenvector.
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